Which of the following is equivalent to the complex number i^(35) ? Choose 1 answer: (A) 1 (B) i (C) -1 (D) -i

Definition of i: To solve for i^(35), we need to remember that i is the imaginary unit, which is defined by i^2 = -1. We can use the powers of i to simplify i^(35) by recognizing the pattern that repeats every four powers: i, -1, -i, 1, and then back to i. Let's find the remainder when 35 is divided by 4 to determine where i^(35) falls in this pattern.Pattern of powers of i: Divide 35 by 4 to find the remainder. 35 divided by 4 is 8 with a remainder of 3. This means that i^(35) is equivalent to i^(4*8+3), which simplifies to (i^4)^8 * i^3. Since i^4 = 1, this further simplifies to 1^8 * i^3.Finding the remainder: Now we calculate 1^8 and i^3. Since any number to the power of 8 is itself, 1^8 is 1. Then we calculate i^3, which is i^2 * i. We know i^2 is -1, so i^3 is -1 * i, which is -i.Simplifying i^(35): Combining the results from the previous steps, we have 1 * -i, which is simply -i. Therefore, i^(35) is equivalent to -i.

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Which of the following is equivalent to the complex number
i^(35) ?
Choose 1 answer:
(A) 1
(B)
i
(C) -1
(D)
-i

Which of the following is equivalent to the complex number $i^{35}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $1$ $\newline$ (B) $i$ $\newline$ (C) $-1$ $\newline$ (D) $-i$

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Math Problems

Algebra 1

Domain and range of quadratic functions: equations

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Q. Which of the following is equivalent to the complex number

i^{35}

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

i

\newline

(C)

-1

\newline

(D)

-i

Definition of i: To solve for $i^{35}$ , we need to remember that $i$ is the imaginary unit, which is defined by $i^2 = -1$ . We can use the powers of $i$ to simplify $i^{35}$ by recognizing the pattern that repeats every four powers: $i$ , $-1$ , $-i$ , $1$ , and then back to $i$ . Let's find the remainder when $i$ $0$ is divided by $i$ $1$ to determine where $i^{35}$ falls in this pattern.
Pattern of powers of i: Divide $35$ by $4$ to find the remainder. $35$ divided by $4$ is $8$ with a remainder of $3$ . This means that $i^{35}$ is equivalent to $i^{4\cdot8+3}$ , which simplifies to $(i^4)^8 \cdot i^3$ . Since $i^4 = 1$ , this further simplifies to $4$ $0$ .
Finding the remainder: Now we calculate $1^8$ and $i^3$ . Since any number to the power of $8$ is itself, $1^8$ is $1$ . Then we calculate $i^3$ , which is $i^2 \cdot i$ . We know $i^2$ is $-1$ , so $i^3$ is $i^3$ $0$ , which is $i^3$ $1$ .
Simplifying $i^{35}$ : Combining the results from the previous steps, we have $1 \cdot (-i)$ , which is simply $-i$ . Therefore, $i^{35}$ is equivalent to $-i$ .